Question

Maximum value of logx/x​

Answer 1

Answer:

1 / x^2

Step-by-step explanation:

lnx /x =1/x^2 ok

follow

Answer 2

Answer:

By Fermat’s interior extremum theorem, the local extrema, which happens to be a global maximum in this case, occurs where f′(x)=0 . It also speaks of boundaries and non-differentiable points though this is irrelevant in our case.

So, let’s differentiate f(x)=log(x)x ,

f′(x)=ddx[1x]log(x)+ddx[log(x)]1x=−log(x)x2+1x1x=1−log(x)x2 .

Let’s work out x where f′(x)=0 .

1−log(x)x2=0⟹1−log(x)=0⟹log(x)=1 . Simply exponentiate both sides to reach x=exp(1)=e .

e∈[2,∞) , so f(e) is the local maximum over [2,∞) . It is also the global maximum for x∈(0,∞) .

Edit: I just noticed that you asked for the maximum value, that would be f(e)=log(e)e=1e .

Step-by-step explanation:

hope it helps you